a companys revenue (in thousands of dollars) from a campaign with budget x (thousands of dollars) is s(x)=…

a companys revenue (in thousands of dollars) from a campaign with budget x (thousands of dollars) is s(x)= - 0.012x^3 + 0.66x^2 + 120, 0≤x≤55. (a) find where s is increasing and decreasing. increasing on: decreasing on: (b) enter the x - value where s has its relative and absolute maximum. (c) determine the intervals of concavity. concave up on: concave down on: (d) find the inflection point (enter the x - value).

a companys revenue (in thousands of dollars) from a campaign with budget x (thousands of dollars) is s(x)= - 0.012x^3 + 0.66x^2 + 120, 0≤x≤55. (a) find where s is increasing and decreasing. increasing on: decreasing on: (b) enter the x - value where s has its relative and absolute maximum. (c) determine the intervals of concavity. concave up on: concave down on: (d) find the inflection point (enter the x - value).

Answer

Explanation:

Step1: Find the first - derivative

The revenue function is $S(x)= - 0.012x^{3}+0.66x^{2}+120$, where $0\leq x\leq55$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $S^\prime(x)=-0.036x^{2}+1.32x$.

Step2: Find the critical points

Set $S^\prime(x) = 0$. So, $-0.036x^{2}+1.32x = 0$. Factor out $x$: $x(-0.036x + 1.32)=0$. This gives $x = 0$ or $-0.036x+1.32 = 0$. Solving $-0.036x+1.32 = 0$ for $x$: $0.036x=1.32$, then $x=\frac{1.32}{0.036}=\frac{1320}{36}=\frac{110}{3}\approx36.7$.

Step3: Determine increasing and decreasing intervals

We use test points in the intervals $[0,\frac{110}{3})$ and $(\frac{110}{3},55]$. For the interval $[0,\frac{110}{3})$, let's choose $x = 10$. Then $S^\prime(10)=-0.036\times10^{2}+1.32\times10=- 3.6 + 13.2 = 9.6>0$. So, $S(x)$ is increasing on $[0,\frac{110}{3})\approx[0,36.7)$. For the interval $(\frac{110}{3},55]$, let's choose $x = 40$. Then $S^\prime(40)=-0.036\times40^{2}+1.32\times40=-0.036\times1600 + 52.8=-57.6+52.8=-4.8<0$. So, $S(x)$ is decreasing on $(\frac{110}{3},55]\approx(36.7,55]$.

Step4: Find the relative and absolute maximum

Since $S(x)$ is increasing on $[0,\frac{110}{3})$ and decreasing on $(\frac{110}{3},55]$, the relative (and absolute since it's the only one in the domain) maximum occurs at $x=\frac{110}{3}\approx36.7$.

Step5: Find the second - derivative

$S^{\prime\prime}(x)=\frac{d}{dx}(-0.036x^{2}+1.32x)=-0.072x + 1.32$.

Step6: Determine concavity

Set $S^{\prime\prime}(x)=0$. Then $-0.072x + 1.32 = 0$, so $0.072x = 1.32$ and $x=\frac{1.32}{0.072}=\frac{1320}{72}=\frac{55}{3}\approx18.3$. For $x<\frac{55}{3}$, let's choose $x = 10$. Then $S^{\prime\prime}(10)=-0.072\times10 + 1.32=-0.72 + 1.32 = 0.6>0$. So, $S(x)$ is concave up on $[0,\frac{55}{3})\approx[0,18.3)$. For $x>\frac{55}{3}$, let's choose $x = 20$. Then $S^{\prime\prime}(20)=-0.072\times20+1.32=-1.44 + 1.32=-0.12<0$. So, $S(x)$ is concave down on $(\frac{55}{3},55]\approx(18.3,55]$.

Step7: Find the inflection point

Set $S^{\prime\prime}(x) = 0$. We found $x=\frac{55}{3}\approx18.3$.

Answer:

(a) Increasing on $[0,\frac{110}{3})\approx[0,36.7)$; Decreasing on $(\frac{110}{3},55]\approx(36.7,55]$ (b) $x=\frac{110}{3}\approx36.7$ (c) Concave up on $[0,\frac{55}{3})\approx[0,18.3)$; Concave down on $(\frac{55}{3},55]\approx(18.3,55]$ (d) $x=\frac{55}{3}\approx18.3$